Answer: B
Remark: for $w,z\in(0,1)$, $w^m< w^n; w^m>z^m$ for $m>n; w>z; m,n\in\mathbb{N}$.
We note that A, B are symmetric sets. Hence, we can consider the values of $x$ and $y$ only in the first quadrant.
Further, we consider the extreme values of the set A and set B i.e.
$$A_1=\{(x,y):x^2+y^4=1\}$$ and $$B_1=\{(x,y):x^4+y^6=1\}$$
Moreover, for fixed $x\in (0,1)$ \begin{align*}x^2>& x^4\\ \Rightarrow-x^2<&-x^4\\\Rightarrow 1-x^2<&1-x^4\\\Rightarrow y_{A_1}^4<&y_{B_1}^6\\\Rightarrow y_{A_1}^4<&y_{B_1}^4 \enspace \text{[see Remark]}\\\Rightarrow y_{A_1}<&y_{B_1}\end{align*}
Thus, $A\subseteq B$.